Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z^2 - 6z - 40}{z + 1} \times \dfrac{z + 1}{z - 10} $
Answer: First factor the quadratic. $x = \dfrac{(z - 10)(z + 4)}{z + 1} \times \dfrac{z + 1}{z - 10} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (z - 10)(z + 4) \times (z + 1) } { (z + 1) \times (z - 10) } $ $x = \dfrac{ (z - 10)(z + 4)(z + 1)}{ (z + 1)(z - 10)} $ Notice that $(z + 1)$ and $(z - 10)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ \cancel{(z - 10)}(z + 4)(z + 1)}{ (z + 1)\cancel{(z - 10)}} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $x = \dfrac{ \cancel{(z - 10)}(z + 4)\cancel{(z + 1)}}{ \cancel{(z + 1)}\cancel{(z - 10)}} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $x = z + 4 ; \space z \neq 10 ; \space z \neq -1 $